Evaluate, in spherical coordinates, the triple integral of*f*(*ρ*,*θ*,*ϕ*)=cos*ϕ*, over the region0≤*θ*≤2*π*,*π*/3≤*ϕ*≤*π*/2,3≤*ρ*≤7integral=

We are given the function \( f(\rho, \theta, \phi) = \cos \phi \) and asked to evaluate the triple integral in spherical coordinates over a specific region:

- \( 0 \leq \theta \leq 2\pi \)
- \( \frac{\pi}{3} \leq \phi \leq \frac{\pi}{2} \)
- \( 3 \leq \rho \leq 7 \)

The volume element in spherical coordinates is \( dV = \rho^2 \sin \phi \, d\rho \, d\theta \, d\phi \).

Thus, the triple integral becomes:

\[
\int_0^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_3^7 (\cos \phi) \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta
\]

### Step-by-step solution:

1. **Integrate with respect to \( \theta \):**

Since the integrand does not depend on \( \theta \), the integral over \( \theta \) is straightforward:

\[
\int_0^{2\pi} d\theta = 2\pi
\]

2. **Integrate with respect to \( \rho \):**

We next integrate with respect to \( \rho \). The integrand involving \( \rho \) is \( \rho^2 \). So we compute:

\[
\int_3^7 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_3^7 = \frac{7^3}{3} - \frac{3^3}{3} = \frac{343}{3} - \frac{27}{3} = \frac{316}{3}
\]

3. **Integrate with respect to \( \phi \):**

The integrand for \( \phi \) is \( \cos \phi \sin \phi \), which can be simplified using the identity \( \cos \phi \sin \phi = \frac{1}{2} \sin(2\phi) \):

\[
\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cos \phi \sin \phi \, d\phi = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sin(2\phi) \, d\phi
\]

Let \( u = 2\phi \), so \( du = 2 d\phi \), and the limits change accordingly:

- When \( \phi = \frac{\pi}{3}, u = \frac{2\pi}{3} \)
- When \( \phi = \frac{\pi}{2}, u = \pi \)

Thus, the integral becomes:

\[
\frac{1}{4} \int_{\frac{2\pi}{3}}^{\pi} \sin u \, du = \frac{1}{4} \left[ -\cos u \right]_{\frac{2\pi}{3}}^{\pi}
\]

Evaluating the cosine terms:

\[
= \frac{1}{4} \left( -\cos(\pi) + \cos\left(\frac{2\pi}{3}\right) \right)
= \frac{1}{4} \left( 1 - \left( -\frac{1}{2} \right) \right)
= \frac{1}{4} \left( 1 + \frac{1}{2} \right) = \frac{1}{4} \times \frac{3}{2} = \frac{3}{8}
\]

4. **Final result:**

Now, multiplying all the results together:

\[
2\pi \times \frac{316}{3} \times \frac{3}{8} = 2\pi \times \frac{316}{8} = \frac{316\pi}{4} = 79\pi
\]

Thus, the value of the triple integral is:

\[
\boxed{79\pi}
\]

Would you like any details or further explanations?

Here are 5 related questions:
1. How does spherical coordinates differ from cylindrical coordinates in volume integrals?
2. What are the volume elements in cylindrical coordinates?
3. How do trigonometric identities simplify integrals?
4. What is the role of Jacobian in transforming between coordinate systems?
5. Can you explain the divergence theorem in the context of spherical coordinates?

**Tip:** For spherical coordinates, always remember to include the volume element \( \rho^2 \sin \phi \) when setting up integrals!

Evaluate, in spherical coordinates, the triple integral of f(ρ,θ,ϕ)=cosϕ, over the region 0≤θ≤2π, π/3≤ϕ≤π/2, 3≤ρ≤7

Learn how to evaluate the triple integral of f(ρ,θ,ϕ)=cosϕ in spherical coordinates over the region 0≤θ≤2π, π/3≤ϕ≤π/2, and 3≤ρ≤7. This solution includes key calculus concepts, such as volume integration in spherical coordinates and the use of trigonometric identities. Suitable for high school students studying calculus.

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